Hi all,
in continuation to
http://tech.groups.yahoo.com/group/fc-solve-discuss/message/1115 ,
I was thinking how to represent Baker’s Dozen positions.
What I came with was:
1. I can use 4 * log2(14) data to represent the foundations.
2. The 13 bottom-most cards will always be either in the foundations or in
their original place (due to the prune at
http://tech.groups.yahoo.com/group/fc-solve-discuss/message/1121 ).
We can tell which one is that based on the encodings in #1 - so:
13 * log2(1)
3. The rest of the cards can be either in their original position or on top of
one of the 4 parents of all suits - H, C, D, S. So 39 * log2(5).
To sum up:
$ perl -MLog2 -e 'plog2(4 => 14, 13 => 1, 39 => 5)'
log2(14) * 4 + log2(1) * 13 + log2(5) * 39 = 105.784615388838 bits (bytes: 14)
so this is even better than Freecell. There may be some complications in trying
to represent games in mid-play where the prune was not completely followed.
For this, there is:
$ perl -MLog2 -e 'plog2(4 => 14, 48 => 5)'
log2(14) * 4 + log2(5) * 48 = 126.681968242824 bits (bytes: 16)
(Because the kings cannot be moved).
Regards,
Shlomi Fish
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Received on Tue Jul 03 2012 - 04:02:36 IDT